What are the solutions to the equation 3w^2-12w+12=0 ?

The solutions to the equation 3w^2-12w+12=0 are w=2

Find the zeros of 3w^2-12w+12=0

The zeros of 3w^2-12w+12=0 are w=2

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3 w^{2} - 12 w + 12 = 0

w = 2

Solution steps

3 w^{2} - 12 w + 12 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 12}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 12 = 0

w_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 3}

w = \frac{- ( - 12 )}{2 \cdot 3}

\frac{- ( - 12 )}{2 \cdot 3} = 2

w = 2

The solution to the quadratic equation is:

w = 2

8 w^{2} + 2 w + 2 = 3

lo g_{9} (3) + lo g_{9} (27)

x^{2} - 3 x - 10 \geq 0

4 m^{2} - 25 = 0

\frac{x y - x - y + 1}{x ^{2} + 9 x - 10}

What are the solutions to the equation 3w^2-12w+12=0 ?
The solutions to the equation 3w^2-12w+12=0 are w=2
Find the zeros of 3w^2-12w+12=0
The zeros of 3w^2-12w+12=0 are w=2